Question
In $\triangle ABC, \angle B = 90^\circ .$ If $AB = 12$ units and $BC = 5$units, find: $\cot C$
✓
Answer
In $\triangle ABC,$
$AC^2 = AB^2 + BC^2$
$\Rightarrow AC =\sqrt{ AB ^2+ BC ^2}$
$\Rightarrow AC =\sqrt{12^2+5^2}$
$=\sqrt{144+25}$
$= 13$
$AB = 12$ units
$BC = 5$ units
$AC = 13$ units
$\cot C$
$=\frac{\text { Base }}{\text { Perpendicular }}$
$=\frac{ BC }{ AB }$
$=\frac{5}{12} .$
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